The formula for the Weyl symbol [1]} implies that with \(\mathcal {U}\)  being the unitary map used in the Proof of Lemma  REF
\(\begin{split}\widehat{\operatorname{tr}}(f(H )) &= \lim _{n \rightarrow \infty } \frac{ \operatorname{tr}_{\ell ^2({\mathbb {Z}})}(\operatorname{1\hspace{-2.75pt}l}_{\lbrace -n,...,n\rbrace } f(H)) }{2n+1} \\&= \lim _{n \rightarrow \infty } \frac{\operatorname{tr}_{\ell ^2({\mathbb {Z}})}(\operatorname{1\hspace{-2.75pt}l}_{\lbrace -n,...,n\rbrace } \mathcal {U}^{-1} f(H_{\operatorname{\Psi DO}}) \mathcal {U})}{2n+1} \\&= \lim _{n \rightarrow \infty } \frac{\sum _{\gamma \in \lbrace -n,..,n\rbrace ^2} \operatorname{tr}_{{\mathbb {C}}^4}( \int _{\mathbb {T}_1} e^{-i \gamma x} f(H_{\operatorname{\Psi DO}}( x,h D_x,\theta )) e^{i \gamma x } \, d x)}{2n+1} \\&= \lim _{n \rightarrow \infty } \frac{\sum _{\gamma \in \lbrace -n,..,n\rbrace ^2} \operatorname{tr}_{{\mathbb {C}}^4}( \int _{\mathbb {T}_1}\sigma (f(H_{\operatorname{\Psi DO}}))(x,\gamma L^{-1}) \, d x)}{2n+1 }.\end{split}\) 
When \(1/L\)  is rational
\(\gamma \mapsto \operatorname{tr}_{{\mathbb {C}}^4}\left( \int _{\mathbb {T}_1}\sigma (f(H_{\operatorname{\Psi DO}}))(x,\gamma L^{-1}) \, d x\right)\) 
is periodic and thus we obtain
\( \widehat{\operatorname{tr}}(f(H )) = \sum _{\gamma \in \lbrace 0,..,q-1\rbrace } \operatorname{tr}_{{\mathbb {C}}^4} \int _{\mathbb {T}_1}\sigma (f(H_{\operatorname{\Psi DO}}))(x,\gamma L^{-1}) \, dx.\) 
If \(1/L\)  does not satisfy this rationality condition, then the translation \((T^{\gamma }u)(x) = u(x+1/L)\)  is a uniquely ergodic endomorphism on the probability space \({\mathbb {R}}/{\mathbb {Z}}\)  and therefore using the continuity of the Weyl symbol, it follows that [2]}
\( \widehat{\operatorname{tr}}(f(H )) =\int _{{\mathbb {R}}^2/{\mathbb {Z}}^2}\operatorname{tr}_{{\mathbb {C}}^4}\sigma (f(H_{\operatorname{\Psi DO}}))(x,\xi ) \, dx \, d\xi .\) 
